How to prove that $(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$ if $(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$

Question

Let $x,y$ be real numbers such that :

$(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$.

Prove that :

$(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$.

I tried taking $x=y$. It simplifies everything a lot. But I'm not able to progress when both $x$ and $y$ are in the same equation.

Answer 1

Let $x =\sinh(a), y=\sinh(b)$.

Then $$(\sinh(a)+\cosh(b)) (\cosh(a) +\sinh(b))=1 \\ \sinh(a)\cosh(a)+\sinh(b)\cosh(b)+\cosh(a)\cosh(b)+\sinh(a)\sinh(b)=1 \\ \sinh(a+b)+\cosh(a+b)=1 $$

Now, $$1=\cosh^2(a+b)-\sinh^2(a+b)=\left( \sinh(a+b)+\cosh(a+b)\right)\left( \cosh(a+b)-\sinh(a+b)\right) \\=\left( \cosh(a+b)-\sinh(a+b)\right)$$ Therefore $$\sinh(a+b)+\cosh(a+b)=1\\ \cosh(a+b)-\sinh(a+b)=1$$ and hence $$\cosh(a+b)=1 \\ \sinh(a+b)=0$$

which proves that $a+b=0$.

Therefore $y=-x$.

From here it is trivial.

Answer 2

HINT:

Write $$x = \frac{1}{2}(s-\frac{1}{s})\\ y = \frac{1}{2}(t-\frac{1}{t})$$ with $s$, $t>0$. Then

$$( x+ \sqrt{1+y^2})(y + \sqrt{1+x^2}) -1 = (s t -1) \frac { s t (s+t)^2 + (s-t)^2}{ 4 s^2 t^2}$$